A parabola passes through the point (0, 12) on the y-axis and intersects the x-axis at two… — SAT

A parabola passes through the point $(0, 12)$ on the $y$ -axis and intersects the $x$ -axis at two distinct points, $P$ and $Q$ . The line connecting $P$ to the $y$ -intercept has a slope of $2$ , while the line connecting $Q$ to the $y$ -intercept has a slope of $-6$ . What is the maximum $y$ -value achieved by this parabola?

Answer: 16

Answer

The maximum

y

-value achieved by the parabola is 16.

The slopes of the lines connecting the

x

-intercepts

P(x_1, 0)

and

Q(x_2, 0)

to the

y

-intercept

C(0, 12)

determine their coordinates. The slope of the line through

P

and

C

\frac{12 - 0}{0 - x_1} = 2

, which gives

x_1 = -6

. The slope of the line through

Q

and

C

\frac{12 - 0}{0 - x_2} = -6

, which gives

x_2 = 2

. The factored form of the parabola is

y = a(x + 6)(x - 2)

. Using the

y

-intercept

C(0, 12)

, we find

12 = a(6)(-2)

, meaning

a = -1

. The

x

-coordinate of the vertex is the midpoint of the intercepts, which is

\frac{-6 + 2}{2} = -2

. Substituting

x = -2

into the equation gives

y = -(-2 + 6)(-2 - 2) = 16

. Since the leading coefficient is negative, this represents the maximum

y

-value.

Step-by-Step Solution

Find the coordinates of

P(x_1, 0)

and

Q(x_2, 0)

using the slopes of the lines from these points to the

y

-intercept

C(0, 12)

P = (-6, 0)

and

Q = (2, 0)

The slope formula between

(x_1, 0)

and

(0, 12)

\frac{12 - 0}{0 - x_1} = 2

, which gives

x_1 = -6

. Similarly, for the second line,

\frac{12 - 0}{0 - x_2} = -6

, which gives

x_2 = 2

Write the quadratic equation in factored form using the

x

-intercepts.

y = a(x + 6)(x - 2)

Since the parabola intersects the

x

-axis at

(-6, 0)

and

(2, 0)

, these points are the roots of the quadratic function.

Solve for the leading coefficient

a

by substituting the

y

-intercept

(0, 12)

into the equation.

a = -1

Substituting

x = 0

and

y = 12

gives

12 = a(0 + 6)(0 - 2)

, which simplifies to

12 = -12a

, so

a = -1

Determine the

x

-coordinate of the vertex.

x = -2

The axis of symmetry of a parabola lies exactly midway between its

x

-intercepts:

x = \frac{-6 + 2}{2} = -2

Calculate the maximum

y

-value by evaluating the function at the

x

-coordinate of the vertex.

y = 16

Substituting

x = -2

into

y = -(x + 6)(x - 2)

gives

y = -(-2 + 6)(-2 - 2) = -(4)(-4) = 16

Key Concept

Using

x

-intercepts and coordinate geometry to find the vertex of a parabola.

Question

Answer

Step-by-Step Solution

Key Concept

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